Chestermiller said:
However, according to russ waters, to observers on the Earth's surface, the clocks on the shuttle are running faster than their own, and it is Earth surface clocks that are running at the usual rate.
I don't think russ's statement was correct for the Space Shuttle. It is correct for satellites in high orbits, such as the GPS satellites, where the effect of altitude (speeding up clocks relative to those on the Earth's surface) outweighs the effect of orbital velocity (slowing down clocks relative to those on the Earth's surface). For satellites in low Earth orbit, such as the Shuttle, the velocity effect outweighs the altitude effect.
An approximate equation for the rate of time flow for an object in Earth's gravity field is
\frac{d\tau}{dt} = \sqrt{1 - \frac{2 G M}{c^{2} r} - \frac{v^{2}}{c^{2}}}
For an object on Earth's surface, we can eliminate the v^2 term by using the poles as a reference; since the Earth's surface is (approximately) an equipotential surface, clocks on the surface all go at the same rate, and the rate at the poles is the easiest to calculate. The polar radius of the Earth is 6.36 x 10^6 meters according to Wikipedia. G is 6.67 x 10^-11, and M for the Earth is 5.97 x 10^24 kg. This gives a result of d\tau / dt = 1 - 6.97 * 10^{-10}.
For objects in orbit, since v^2 = GM/r for an object in a free-fall circular orbit (which we'll assume is a good enough approximation for this problem), we can rewrite the above more simply as
\frac{d\tau}{dt} = \sqrt{1 - \frac{3G M}{c^{2} r}}
For the Space Shuttle, r is about 200 miles, or 3.2 x 10^5 m, above the Earth's surface. For the GPS satellites, r is about 4.2 Earth radii. The corresponding numbers are:
Shuttle: d\tau / dt = 1 - 9.95* 10^{-10}, which is slower than ground clocks by about 3 parts in 10^-10.
GPS satellite: d\tau / dt = 1 - 2.49 * 10^{-10}, which is faster than ground clocks by about 4.5 parts in 10^-10.
